MATH 100, Section 110 (CSP)
Week 7: Marked Homework Solutions
2010 Oct 28
1. [10] In the figure below, y(t) is the height of the ball in m, x(t) is the distance (also in m)
between the ball’s shadow and the point on the ground directly under the ball. The units
for time t are s.
(a) Given: y 0 . Required: x0 . So we need to relate y and x. By similar triangles,
x
x+3
=
,
y
8
and differentiating this expression with respect to t gives
x0 y − y 0 x
x0
=
.
y2
8
(1)
At the instant when the ball is 4 m above the ground, we use similar triangles to get
x = 3y/(8 − y) and
y = 4,
x = 3,
y 0 = −20.
Substituting these values into equation (1) and solving for x0 , we get
x0 (4) − (−20)(3)
x0
= ,
16
8
x0
60
=− ,
8
16
x0 = −30.
The shadow is moving 30 m/s (towards the point directly below the ball) when the ball is 4
m above the ground.
(b) Let z(t) be the distance (in m) between the ball and its shadow. By Pythogoras’ Theorem,
z 2 = x2 + y 2 ,
1
and differentiating this with respect to t gives
2zz 0 = 2xx0 + 2yy 0 .
At
p the instant
√ when the ball is 4 m above the ground, Pythogoras’ Theorem gives z =
2
2
x + y = 9 + 16 = 5 and
5z 0 = (3)(−30) + (4)(−20) = −170,
so z 0 = −170/5 = −34. When the ball is 4 m above the ground, the distance between the
ball and its shadow is decreasing at 34 m/s.
2. [10] We require the rate at which water is being pumped in Vin0 , where V (t) is the total
volume of water in the tank, in cm3 . Let h(t) be the water level, in cm, with time t measured
in min.
0
We know the rate at which water is leaking out Vout
= 10, 000 cm3 /min, and the rate the
water level is changing is h0 = 20 cm/min at the instant when h = 200 cm. If we find the
total rate of volume change of water in the tank V 0 in cm3 /min, then
0
V 0 = Vin0 − Vout
,
and solving for the required rate gives
0
Vin0 = V 0 + Vout
= V 0 + 10, 000.
We can find V 0 in terms of h0 , because there is a relation between V and h. The volume V
of water is the volume of the cone
1
V = πr 2 h.
3
But we have to eliminate r, since we are not given the corresponding rate. By similar
, so
triangles, hr = 200
600
r = 13 h
and therefore
1 3
πh ,
27
a general expression valid not only just when h = 200 cm. Differentiating this last expression
gives
1
V 0 = πh2 · h0 .
9
V =
2
Evaluating at the instant when h = 200 cm, we get V 0 = 19 π(200)2 · 20 =
Therefore the rate at which water is being pumped in is
Vin0 = V 0 + 10, 000 =
800,000π
9
cm3 /min.
800, 000π
+ 10, 000 (≈ 289, 000) cm3 /min.
9
3. [10] In the figure below showing the cross-section of the cylinder and cone, H(t) is the
height of the water in the cylinder (H ≥ 7), y(t) is the distance between the bottom of the
cylinder and the tip of the cone (both distances in cm). Time t is measured in s. When the
tip of the cone is below the top surface of the water, 0 ≤ y ≤ 7 and the part of the cone that
is under water is itself a cone of height H − y and base radius r.
We are given y 0 = −1 cm/s and H 0 is required. To get a relation between H and y, we use
the fact that the volume of water is constant. Before the cone is lowered into the water, the
volume is π52 7 = 175π cm3 . When the tip of the cone is below the surface of the water,
assuming that the cone is nonporous and displaces a volume of water equal to the volume
of the submerged part of the cone, the volume of the water is
[Volume of a cylinder of radius 5 and height H]
−[Volume of a cone of height (H − y) and base radius r] = 175π.
By similar triangles,
r
3
= ,
H −y
4
r=
3(H − y)
,
4
so the equation for the volume of water becomes
2
1 3(H − y)
(H − y) = 175π
25πH − π
3
4
or
25H −
3
(H − y)3 = 175.
16
Differentiating with respect to t gives
25H 0 −
9
(H − y)2 (H 0 − y 0 ) = 0.
16
3
At the moment the cone is completely submerged, H − y = 4 and y 0 = −1. Substituting
these values and solving for H 0 gives
H 0 = 9/16,
the water level is rising at 9/16 cm/s.
4. [3] (a) Let f (x) = ln x and let a = 1. Then f 0 (x) = 1/x, f 0 (1) = 1 and the linearization
of f at 1 is L(x) = f (1) + f 0 (1)(x − 1) = ln 1 + (x − 1) = x − 1, L(0.98) = −0.02. So we
estimate
ln 0.98 ≈ −0.02.
(The actual value is −0.020203 accurate to six decimal places.)
[3] (b) For the formulas we have derived using calculus, angles must be in radians. So 46o
is 46π/180 radians. Let f (x) = tan x and let a = π/4 (or 45o ). Then f 0 (x) = sec2 x,
f 0 (π/4) = sec2 (π/4) = 2, and the linearization of f at π/4 is L(x) = f (π/4) + f 0 (π/4)(x −
π/4) = 1 + 2(x − π/4), L(46π/180) = 1 + 2(π/180) = 1 + (π/90) ≈ 1.0349. So we estimate
tan 46◦ = 1 +
π
(≈ 1.0349).
90
(The actual value is 1.035530 accurate to six decimal places.)
5. [3] We are given that ∆r = dr = 0.05r, and linear approximation for the corresponding change in F is ∆F ≈ dF , where ∆F is the actual change in F , and dF = 4kr 3 dr =
4kr 3 (0.05r) = 0.20kr 4 is the linear approximation to the change in F . The model predicts
∆F/F ≈ dF/F = 0.20kr 4 /kr 4 = 0.20, so the flux of blood would increase by approximately
20%, assuming that ∆r is small enough for the linear approximation to be sufficiently accurate. Since ∆F is positive, the flux increases.
6. (Check that x = 1, y = 1 in fact is a solution.) [4] (a) Linear approximation at x = 1 is
L(x) = T1 (x) = f (1) + f 0 (1)(x − 1),
where f (1) = 1 is given. Use implicit differentiation to find f 0 (1):
4y 3 y 0 + 3y 2 y 0 + y + xy 0 = 0
gives y 0 =
−y
,
4y 3 +3y 2 +x
so when (x, y) = (1, 1) we have
y 0 = f 0 (1) =
−1
= − 18 .
4+3+1
Therefore the linear approximation is
L(x) = 1 − 18 (x − 1),
and
y ≈ L(0.9) = 1.0125
4
when x = 0.9, using linear approximation.
[6] (b) The quadratic approximation at x = 1 is
Q(x) = T2 (x) = f (1) + f 0 (1)(x − 1) + 2!1 f 00 (1)(x − 1)2 ,
with f (1) = 1 given and f 0 (1) = 81 calculated above. To find f 00 (1), use implicit differentiation
again:
12y 2 y 0 · y 0 + 4y 3 · y 00 + 6yy 0 · y 0 + 3y 2 · y 00 + y 0 + 1 · y 0 + x · y 00 = 0
(or differentiate the expression for y 0 using the quotient rule to get y 00 =
Then substitute
x = 1, y = 1, y 0 = − 18
−y 0 (4y 3 +3y 2 +x)+y(12y 2 y 0 +6yy 0 +1)
).
(4y 3 +3y 2 +x)2
and obtain
1
y 00 = f 00 (1) = − 256
.
Therefore the quadratic approximation is
T2 (x) = 1 − 18 (x − 1) −
1
(x
512
− 1)2 ,
and using this we get
y ≈ T2 (0.9) = 1 − 81 (−0.1) −
1
(−0.1)2
512
= 1.012480
when x = 0.9.
7. [3]
f (x) = x−1/2 , f 0 (x) = − 12 x−3/2 , f 00 (x) = 43 x−5/2 ,
1
f (9) = 9−1/2 = 13 , f 0 (9) = − 21 · 9−3/2 = − 54
, f 00 (9) =
T2 (x) =
1
3
−
1
(x
54
− 9) +
3
4
1
(x
648
· 9−5/2 =
1
,
324
− 9)2 .
8. [3]
f (x) = cos x, f (1) (x) = − sin x, f (2) (x) = − cos x, f (3) (x) = sin x, f (4) (x) = cos x, etc.
f (0) = 1, f (1) (0) = 0, f (2) (0) = −1, f (3) (0) = 0, f (4) (0) = 1, etc.
T8 (x) = 1 − 2!1 x2 + 4!1 x4 − 6!1 x6 + 8!1 x8 .
5